1) Let O be the center of the circle. Then the triangle ODA and ODC are congruent (because OD=OD, OA=OC and angles ODA and ODC are right). Therefore AD=DC.
Then the height BD is also median and that happens in an isosceles triangle.
2)
AC also bisects the arc BD. Then $\displaystyle \widehat{BEC}=\widehat{CED}$ But $\displaystyle \widehat{BEC}+\widehat{CED}=180\Rightarrow \widehat{BEC}=\widehat{CED}=90$